This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.
Divide largest force.
Forces will equalize themselves.
The bearing of these heads against the pin we know are all right, also the shearing, as the greatest shearing under this arrangement will be a single shear, between E and C (or E111 and C1) and equal 22000 pounds, or considerably less than half the safe single shearing <m this pin, which we previously found to be 48000 pounds.
The bending-moments on the pin will now be, at C: (Left side) mc = 22000.27/32 - 32000.0 = 18563 pounds-inch. Check (use right side) mc = 22000. (27+73+127/32) - 24000.50/32 - 32000.100/32
= 18563 pounds-inch. A E, we should have : ( Left side) mEl = 22000.50/32 - 32000.27/32 = 10125 pounds-inch. Check (use right side) mE1 = 22000. (46+100/32) - 24000.23/32 - 32000.73/32 = 10125 pounds-inch.
At B we should have : (Either side) mB = 22000. (23+77/32) - 32000.50/32
= 18750 pounds-inch. At the ends, of course, there would be no bending-moment, for take end E we should have : mE = 22000. (154+100 + 54/32) - 32000. (127 + 27/32 ) - 24000.77/32
This arrangement (Figure 189) is therefore satisfactory, so far at least as the strains along the line A E are concerned.
We must now see if it will answer as well for the strains along the other lines (See Figure 182). The direction we shall now have to fear most, will be along the line D A for force D must be placed entirely on the outside edge of pin (not having been located yet) and being quite large, 20000 pounds, may make us some trouble. In Figure 190 we have drawn the forces arranged the same as we settled on last (in Figure 189) but have added the two forces on the extreme ends D and D1 each = DA/2 = 10000 pounds. It will be noticed that along this line (DA Figure 182) the forces E and Care in the same direction. We have divided D into two parts for two reasons. Had we placed it entirely to one side, say to the right of E111, the distance E111 D1, would have been one-quarter of 5/8" larger or = 21/32" ; the leverage of D1 therefore at C1 would have been 27+21/32 = 1 1/2" and the bendingmoment mCl=l 1/2. 20000 = 30000 pounds-inch, too much for our pin.
Examine along other lines.
Besides were we to calculate C and C, by FormulŠ (118 and 119) we should find all strain on C removed and C1 more than doubled. This evidently would not do, without special provision to meet the unequal strain by increasing C1, which would lengthen the pin, so that we prefer to divide D A, making each head of half the thickness, or 5/16" thick. The largest shearing will be between D and C (or D, and C,) and equal 10000 pounds single cross-shear or about 1/5 only of the safe resistance to shearing.
The bending-moments will be (Figure 190) at C. (Left side) mc = 43/32.10000 - 18000.0 = 13437 pounds-inch. Check (right side).
mc = 143/32.10000 + 50/32 .16000 - 100/32.18000 = 13437 pounds-inch, and at B (Either side) mB = 93/32 .10000 - 50/32.18000 = 9375 pounds-inch.
The bearings, of course, are safe, as the thickness of head was originally determined by the largest strain on each rod along its own line. So thai we are all right with. our pin tor strains along line D A. We now take up the strains on the pin along the line A B, Figure 182 which will be as shown in Figure 191.
The greatest Bhearing-strain here will be caused by B and will be a double shear of 28000 pounds, or 14000 pounds on each area, perfectly safe on our size of pin (2 3/4") The moments will be:
(Left side) mE = 1/2.5500 -18375.0
= 2750 pounds-inch. Check (right side) mE = 5500.170/32 + 17250. (27+127/32) + 28000.77/32 -
18375. (54 + 100 + 154/32)
= 2750 pounds-inch. At C:
(Left side) mc = 183 75.27/32 - 5500.43/32
= 8113 pounds-inch.
Check (right side) mc = 18375. (27+73+127/32) - 5500.143/32 - 17250.100/32 -
= 8113 pounds-inch.
(Left side) mEl= 18375.54/32 -5500.70/32- 17250.27/32 = 4422 pounds-inch. Check (right side) mE1 = 18375.(46+100/32) - 28000.23/32- 17250.73/32 - 5500.116/32 = 4422 pounds-inch.
At B: (Either side) mB =18375.(23+77/32) - 17250.50/32 - 5500.93/32 = 14484 pounds-inch.
bo that the arrangement of heads along pin is all right so tar as strains along line A B (Figure 182) are concerned.
We finally examine for the strains along line A C and have strains on the pin accordingly as shown in Figure 192. The greatest shearing here will be between E and C (or C, and E111) and will be
= 4250+18875 = 23125 pounds, single shear, still, less than one-half of the safe single-shear on the pin.
By calculation the moments will be found to be at the different points, as follows :
So that this arrangement of the different bars and strut along the pin is in every way satisfactory.
The graphical method of obtaining bending-moments is frequently much more simple than the arithmetical method ; in important calculations both should be used so as to check each other.
All the rules for formulae given in Chapter VII (Graphical Analysis Of Transverse Strains) for the calculation of transverse strains by the graphical method will apply equally well for pins ; the only difference will be that where there are more than two forces on each side of the pin, the base line of the polygonal Figure between reactions p and q will no longer be straight, but will be composed of several lines.